1.26 g of a dibasic acid were dissolved in water and the solution made up to 200 mL. 20 mL of this solution were completely neutralised by 10 mL of `N/5` NaOH solution. Calculate the equivalent mass and molecular mass of the acid.

Answer

To solve the problem, we need to calculate the equivalent mass and molecular mass of the dibasic acid using the given data. Here’s a step-by-step solution:

### Step 1: Calculate the Molarity of the Acid Solution
We have 1.26 g of dibasic acid dissolved in 200 mL of solution.

**Molarity (M)** is calculated using the formula:
\[ M = \frac{\text{mass of solute (g)}}{\text{molar mass (g/mol)} \times \text{volume of solution (L)}} \]

Converting 200 mL to liters:
\[ 200 \, \text{mL} = 0.200 \, \text{L} \]

So, the molarity of the solution is:
\[ M = \frac{1.26 \, \text{g}}{x \, \text{g/mol} \times 0.200 \, \text{L}} = \frac{1.26}{0.200x} = \frac{6.3}{x} \, \text{mol/L} \]

### Step 2: Calculate the Normality of the Acid
Since the acid is dibasic, it can donate 2 protons (H⁺ ions). The normality (N) of the acid can be calculated using the relation:
\[ N = M \times \text{basicity} \]
Where basicity = 2 for a dibasic acid.

Thus, the normality of the acid is:
\[ N = \frac{6.3}{x} \times 2 = \frac{12.6}{x} \, \text{N} \]

### Step 3: Use the Neutralization Reaction
We know that 20 mL of the acid solution is neutralized by 10 mL of N/5 NaOH solution. First, we convert N/5 to normality:
\[ \text{Normality of NaOH} = \frac{1}{5} \, \text{N} = 0.2 \, \text{N} \]

Using the formula for neutralization:
\[ N_1 V_1 = N_2 V_2 \]
Where:
- \(N_1 = \frac{12.6}{x}\) (normality of the acid)
- \(V_1 = 20 \, \text{mL} = 0.020 \, \text{L}\)
- \(N_2 = 0.2 \, \text{N}\) (normality of NaOH)
- \(V_2 = 10 \, \text{mL} = 0.010 \, \text{L}\)

Substituting the values into the equation:
\[ \frac{12.6}{x} \times 0.020 = 0.2 \times 0.010 \]

### Step 4: Solve for x
Now, simplifying the equation:
\[ \frac{12.6 \times 0.020}{x} = 0.002 \]
Cross-multiplying gives:
\[ 12.6 \times 0.020 = 0.002x \]
\[ 0.252 = 0.002x \]
\[ x = \frac{0.252}{0.002} = 126 \, \text{g/mol} \]

### Step 5: Calculate the Equivalent Mass
The equivalent mass of the dibasic acid can be calculated using the formula:
\[ \text{Equivalent mass} = \frac{\text{Molar mass}}{\text{Basicity}} \]
Since the basicity is 2:
\[ \text{Equivalent mass} = \frac{126 \, \text{g/mol}}{2} = 63 \, \text{g/equiv} \]

### Final Results
- Molecular mass of the dibasic acid = 126 g/mol
- Equivalent mass of the dibasic acid = 63 g/equiv

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